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The compound $\ce{(CH3)2CHCH2NH2}$ can be synthesised by the following route.

$$\ce{CH3CH=CH2 ->[1] CH3CHBrCH3 ->[2] (CH3)2CHCN ->[3] (CH3)2CHCH2NH2}$$

What types of reaction are used in stages 1, 2 and 3?

  • A 1=substitution; 2=addition; 3=reduction

  • B 1=substitution; 2=addition; 3=hydrolysis

  • C 1=addition; 2=substitution; 3=reduction

  • D 1=addition; 2=addition; 3=reduction

  • E 1=addition; 2=substitution; 3=hydrolysis

Could someone please explain to me how they worked out what reactions were used in the stages? I have been able to figure out stage 1 and 2, but I am really stuck on stage 3, I think it should be addition but this isn't one of the answers.

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    $\begingroup$ Well, I had an organic chemistry course more than a decade ago, but I think, in some sense, what happens on the third stage can formally be called an addition reaction. However, it is a very important type of reaction in organic chemistry which usually comes under it is own name. $\endgroup$ – Wildcat Sep 12 '16 at 19:01
  • $\begingroup$ If you correctly identified types of reactions 1 & 2, then you have a simple choice between reduction & hydrolysis for reaction 3. Clearly, only one of these reaction does what is expected. $\endgroup$ – Wildcat Sep 12 '16 at 19:03
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    $\begingroup$ Have you identified the functional groups of the concerned compounds? $\endgroup$ – Loong Sep 12 '16 at 19:28
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C) 1-Addition, 2-Substitution, 3-reduction.

1) The hydrogen bromide is becoming an "addition" to the starting reagent.

2) The cyano functional group "substitutes" for the bromine group.

3) In this reaction the substituted cyano group is reduced to an amine.

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  • $\begingroup$ Oh thank you! I didnt realise that this was a reaction that I needed to learn, I assumed they would have all been able to be worked out! $\endgroup$ – Serra Sep 12 '16 at 20:17
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Most reactions can be classified in a number of different ways, just like many compounds can be classified in a number of different groups. For each transformation, your question only allows for two choices, so you either work out which one works or you just need to identify the wrong choice.

  1. The first reaction is to be classified as either a substitution or an addition. If we look at the structures of reactant and product, we realise that all the reactant’s atoms are present in the product — but an additional $\ce{HBr}$ have been added.

    That should give you more than enough clues that it has to be an addition.

  2. The second reaction is to be classified as either an addition or a substitution. Again, just by checking the atoms present in reactant and product, we realise that we lost the bromine but gained a nitrile group.

    ‘Loss and gain’ is definitely not an addition but a substitution.

  3. The third reaction is to be classified as either a hydrolysis or a reduction. It does get a bit more tricky here. To identify a reduction, you would need to do one of the following:

    • allocate oxidation states and determine whether they have changed
    • check if formally an oxygen has been removed
    • check if formally $\ce{H2}$ has been added.

    Identifying a hydrolysis is easier. It is a reaction with water that can occur in one of two ways:

    • the original molecule is split into two fragments, one of which contains a hydrogen, the other a hydroxy group
    • the molecule has formally added $\ce{H2O}$

    Just by counting carbon atoms, we realise that the molecule has not fragmented, and by identifying atom symbols we can see that water has not formally been added. Thus, it cannot be a hydrolysis, it must be a reduction. In fact, two of the three reduction identification conditions are met, but I’ll leave the proof up to you.

Note that reduction is not the only way in which you can classify reaction 3. You could also call it:

  • hydrogenation
  • hydrogen addition
  • reductive nitrile cleavage
  • saturation
  • and probably others.
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    $\begingroup$ Step 3 is definitely not reductive amination. Cleavage usually refers to sigma bonds breaking, so reductive nitrile cleavage doesn't seem right either. $\endgroup$ – jerepierre Sep 12 '16 at 23:58

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